oh I messed up, the factorials should be 1 less, I skipped n=1 factorial...\[f(x)=\sum_{n=0}^\infty{f^{(n)}(a)\over n!}(x-a)^n\]\[=\frac1{1!}(x-\frac\pi2)-\frac1{3!}(x-\frac\pi2)^3+\frac1{5!}(x-\frac\pi2)^5-...\]

TuringTest

5 years ago

so where does the pi go ?
(no pun intended :P)

anonymous

5 years ago

give me a second to look it this...I'm kinda new and slow at this.

TuringTest

5 years ago

oh no I've messed up terribly, this is all wrong...
I should probably have just slept :P
lol

anonymous

5 years ago

No it's not that. I guess the x and a's and n confuse me.

TuringTest

5 years ago

The formula for the Taylor expansion around \(x=a\) is\[f(x)=\sum_{n=0}^\infty{f^{(n)}(a)\over n!}(x-a)^n\]in your case \(a=\frac\pi2\)
let's go check each term:\[\{a_0\}={f^{(0)}(\frac\pi2)\over0
!}(x-\frac\pi2)^0=\sin\frac\pi2=1\]\[\{a_1\}=\frac1{1!}\cancel{\cos\frac\pi2}^{\huge0}(x-\frac\pi2)^1=0\]\[\{a_2\}=-\frac1{2!}\sin\frac\pi2(x-\frac\pi2)^2=-\frac1{2!}(x-\frac\pi2)^2\]\[\{a_3\}=-\frac1{3!}\cancel{\cos\frac\pi2}^{\huge0}(x-\frac\pi2)^3=0\]so the pattern is that all odd n terms stay

anonymous

5 years ago

Oh darn! I get it now. For a second I forgot what \[f^{(0)}\] and \[f^{(1)}\] meant. Durrr
so the zero derivative is sin(pi/2) and \[(x-\frac{\pi}{2})^0 =1\]
makes sense

anonymous

5 years ago

I get the first line....now on to the second line

TuringTest

5 years ago

one sec, I have to deal with a potentially problematic user...sorry brb

*another typo, the last one should be\[\{a_2\}=-\frac1{2!}(x-\frac\pi2)^2\](I forgot the n=2 in the exponent on the parentheses)

TuringTest

5 years ago

\[\{a_4\}={f^{(4)}(\frac\pi2)\over4!}(x-\frac\pi2)^4\]and\[f(x)=\sin x\implies f^{(4)}(x)=\sin x\implies f^{(4)}(\frac\pi2)=1\]so\[\{a_4\}={1\over4!}(1)(x-\frac\pi2)^4={1\over4!}(x-\frac\pi2)^4\]by now a pattern should be emerging

anonymous

5 years ago

yep I see the pattern now.
Thanks @TuringTest

TuringTest

5 years ago

welcome :)

experimentX

5 years ago

my suggestion ... keep it easy as much as possible
http://www.wolframalpha.com/input/?i=expand+sin%28x%29+at+pi%2F2&dataset=&equal=Submit
http://www.wolframalpha.com/input/?i=expand+cos+x+at+0